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Expand using product rule:

$$P(X \ge a_0 \cap Y \le b_0) = P(X \ge a_0 | Y \le b_0) P(Y \le b_0)$$

Assuming $P(Y \le b_0)$ is nonzero, you can use the first inequality to obtain:

$$P(X \ge a_0 | Y \le b_0) P(Y \le b_0) \le f(Y\le b_0) P(Y \le b_0)$$$$P(X \ge a_0 | Y \le b_0) P(Y \le b_0) \le f(y\le b_0) P(Y \le b_0)$$

Because $f(y)$ is increasing in $y$, $f(Y \le b_0) \le f(b_0)$$f(y \le b_0) \le f(b_0)$ so the inequality of interest holds.

By the way, in the context of probability $f$ is normally used for PDFs so you may want to use a different letter to avoid confusion.

EDIT: Granted, $f(y \le b_0)$ is an abuse of notation. It should be read as $f(y), y \le b_0$ and not as the function taking a set for its argument.

Expand using product rule:

$$P(X \ge a_0 \cap Y \le b_0) = P(X \ge a_0 | Y \le b_0) P(Y \le b_0)$$

Assuming $P(Y \le b_0)$ is nonzero, you can use the first inequality to obtain:

$$P(X \ge a_0 | Y \le b_0) P(Y \le b_0) \le f(Y\le b_0) P(Y \le b_0)$$

Because $f(y)$ is increasing in $y$, $f(Y \le b_0) \le f(b_0)$ so the inequality of interest holds.

By the way, in the context of probability $f$ is normally used for PDFs so you may want to use a different letter to avoid confusion.

Expand using product rule:

$$P(X \ge a_0 \cap Y \le b_0) = P(X \ge a_0 | Y \le b_0) P(Y \le b_0)$$

Assuming $P(Y \le b_0)$ is nonzero, you can use the first inequality to obtain:

$$P(X \ge a_0 | Y \le b_0) P(Y \le b_0) \le f(y\le b_0) P(Y \le b_0)$$

Because $f(y)$ is increasing in $y$, $f(y \le b_0) \le f(b_0)$ so the inequality of interest holds.

By the way, in the context of probability $f$ is normally used for PDFs so you may want to use a different letter to avoid confusion.

EDIT: Granted, $f(y \le b_0)$ is an abuse of notation. It should be read as $f(y), y \le b_0$ and not as the function taking a set for its argument.

Expand using product rule:

$$P(x \ge a_0 \cap y \le b_0) = P(x \ge a_0 | y \le b_0) P(y \le b_0)$$$$P(X \ge a_0 \cap Y \le b_0) = P(X \ge a_0 | Y \le b_0) P(Y \le b_0)$$

Assuming $P(y \le b_0)$$P(Y \le b_0)$ is nonzero, you can use the first inequality to obtain:

$$P(x \ge a_0 | y \le b_0) P(y \le b_0) \le f(y\le b_0) P(y \le b_0)$$$$P(X \ge a_0 | Y \le b_0) P(Y \le b_0) \le f(Y\le b_0) P(Y \le b_0)$$

Because $f(y)$ is increasing in $y$, $f(y \le b_0) \le f(b_0)$$f(Y \le b_0) \le f(b_0)$ so the inequality of interest holds.

By the way, in the context of probability $f$ is normally used for PDFs so you may want to use a different letter to avoid confusion.

Expand using product rule:

$$P(x \ge a_0 \cap y \le b_0) = P(x \ge a_0 | y \le b_0) P(y \le b_0)$$

Assuming $P(y \le b_0)$ is nonzero, you can use the first inequality to obtain:

$$P(x \ge a_0 | y \le b_0) P(y \le b_0) \le f(y\le b_0) P(y \le b_0)$$

Because $f(y)$ is increasing in $y$, $f(y \le b_0) \le f(b_0)$ so the inequality of interest holds.

By the way, in the context of probability $f$ is normally used for PDFs so you may want to use a different letter to avoid confusion.

Expand using product rule:

$$P(X \ge a_0 \cap Y \le b_0) = P(X \ge a_0 | Y \le b_0) P(Y \le b_0)$$

Assuming $P(Y \le b_0)$ is nonzero, you can use the first inequality to obtain:

$$P(X \ge a_0 | Y \le b_0) P(Y \le b_0) \le f(Y\le b_0) P(Y \le b_0)$$

Because $f(y)$ is increasing in $y$, $f(Y \le b_0) \le f(b_0)$ so the inequality of interest holds.

By the way, in the context of probability $f$ is normally used for PDFs so you may want to use a different letter to avoid confusion.

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Expand using product rule:

$$P(x \ge a_0 \cap y \le b_0) = P(x \ge a_0 | y \le b_0) P(y \le b_0)$$

Assuming $P(y \le b_0)$ is nonzero, you can use the first inequality to obtain:

$$P(x \ge a_0 | y \le b_0) P(y \le b_0) \le f(y\le b_0) P(y \le b_0)$$

Because $f(y)$ is increasing in $y$, $f(y \le b_0) \le f(b_0)$ so the inequality of interest holds.

By the way, in the context of probability $f$ is normally used for PDFs so you may want to use a different letter to avoid confusion.